Trigonometry Examples

tan (π + x) = tan (x)

$\tan (π + x) = \tan (x)$

Step 1

Start on the left side.

tan (π + x)

$\tan (π + x)$

Step 2

Apply the sum of angles identity.

\frac{tan (π) + tan (x)}{1 - tan (π) tan (x)}

$\frac{\tan (π) + \tan (x)}{1 - \tan (π) \tan (x)}$

Step 3

Simplify the expression.

Tap for more steps...

Step 3.1

Simplify the numerator.

Tap for more steps...

Step 3.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

\frac{- tan (0) + tan (x)}{1 - tan (π) tan (x)}

$\frac{- \tan (0) + \tan (x)}{1 - \tan (π) \tan (x)}$

Step 3.1.2

The exact value of

tan (0)

$\tan (0)$ is

0

$0$ .

\frac{- 0 + tan (x)}{1 - tan (π) tan (x)}

$\frac{- 0 + \tan (x)}{1 - \tan (π) \tan (x)}$

Step 3.1.3

Multiply

- 1

$- 1$ by

0

$0$ .

\frac{0 + tan (x)}{1 - tan (π) tan (x)}

$\frac{0 + \tan (x)}{1 - \tan (π) \tan (x)}$

Step 3.1.4

Add

0

$0$ and

tan (x)

$\tan (x)$ .

\frac{tan (x)}{1 - tan (π) tan (x)}

$\frac{\tan (x)}{1 - \tan (π) \tan (x)}$

\frac{tan (x)}{1 - tan (π) tan (x)}

$\frac{\tan (x)}{1 - \tan (π) \tan (x)}$

Step 3.2

Simplify the denominator.

Tap for more steps...

Step 3.2.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

\frac{tan (x)}{1 - - tan (0) tan (x)}

$\frac{\tan (x)}{1 - - \tan (0) \tan (x)}$

Step 3.2.2

The exact value of

tan (0)

$\tan (0)$ is

0

$0$ .

\frac{tan (x)}{1 - - 0 tan (x)}

$\frac{\tan (x)}{1 - - 0 \tan (x)}$

Step 3.2.3

Multiply

- - 0

$- - 0$ .

Tap for more steps...

Step 3.2.3.1

Multiply

- 1

$- 1$ by

0

$0$ .

\frac{tan (x)}{1 - 0 tan (x)}

$\frac{\tan (x)}{1 - 0 \tan (x)}$

Step 3.2.3.2

Multiply

- 1

$- 1$ by

0

$0$ .

\frac{tan (x)}{1 + 0 tan (x)}

$\frac{\tan (x)}{1 + 0 \tan (x)}$

\frac{tan (x)}{1 + 0 tan (x)}

$\frac{\tan (x)}{1 + 0 \tan (x)}$

Step 3.2.4

Multiply

0

$0$ by

tan (x)

$\tan (x)$ .

\frac{tan (x)}{1 + 0}

$\frac{\tan (x)}{1 + 0}$

Step 3.2.5

Add

1

$1$ and

0

$0$ .

\frac{tan (x)}{1}

$\frac{\tan (x)}{1}$

\frac{tan (x)}{1}

$\frac{\tan (x)}{1}$

Step 3.3

Divide

tan (x)

$\tan (x)$ by

1

$1$ .

tan (x)

$\tan (x)$

tan (x)

$\tan (x)$

Step 4

Because the two sides have been shown to be equivalent, the equation is an identity.

tan (π + x) = tan (x)

$\tan (π + x) = \tan (x)$ is an identity